As a lot of fluorescent proteins and luciferase existing, an evaluation model to evaluate the performance of every fluorescent protein was established to find the most suitable one.

In this model, Stokes shift, quantity yield (QY), brightness, bleaching time, maturing time and acidity and alkalinity were mainly considered. First, with the analytic hierarchy process, the weight of the above index was obtained. Then, an evaluation matrix was obtained comparing the effect of any two factors ai and aj according to the standard below.

relationship	ai=aj	ai>aj(a little)	ai>aj	ai>aj(obviously)	ai>>aj
score	1	3	5	7	9

There were five factors we should consider. To determine the weight of them, we discussed with Prof. Zhao Kun, school of chemical engineering and technology, Tianjin University and professors in Tianjin Institute of Industrial Biotechnology, Chinese Academy of Sciences, they gave us a suggested order as shown below.

Between every two standards, select numerical value 1, 3, 5, 7 to measure the relative relation between index.

	Stokes	QY*Brightness	Bleaching time	ph	Maturing time
Stokes	1	3	5	1/3	5
QY*Brightness	1/3	1	3	1/5	3
Bleaching time	1/5	1/3	1	1/7	1
ph	3	5	7	1	7
Maturing time	1/5	1/3	1	1/7	1

Construct standard matrix shown as below.

$$A = \begin{bmatrix} 1 & 3 & 5 & 1/3 & 5 \\ 1/3 & 1 & 3 & 1/5 & 3 \\ 1/5 & 1/3 & 1 & 1/7 & 1 \\ 3 & 5 & 7 & 1 & 7 \\ 1/5 & 1/3 & 1 & 1/7 & 1 \end{bmatrix}$$

Next, calculate the concordance index (CI) to judge whether the matrix is reasonable.

$$CI = {\lambda_{max}-n \over n-1} = 0.0340$$

$$\lambda_{max} = 5.1361$$

λ_max -- the largest eigenvalue of the evaluation matrix and n is the dimension of the matrix.

According to the dimension of the matrix, RI value could be found from the form below.

n	1	2	3	4	5	6	7	8	9
RI	0	0	0.58	0.90	1.12	1.24	1.32	1.41	1.45

In our model, because n=5, RI=1.12.

$$CR = {CI \over RI} = 0.0304$$

Because CI<0.1, the matrix passes the consistency check. Therefore, the method we used to find the weight above is suitable.

Calculate the corresponding eigenvector of λ_max and normalize it, W=[0.0762 0.1607 0.3621 0.0389 0.3621].

At last, using the method mentioned in the [1], the score table is shown as below.

Score table
score	0	25	50	75	100
Stokes	0	0-0.02	0.02-0.05	0.05-0.1	0.1-
QY*Brightness	0	0-0.01	0.01-0.04	0.04-0.06	0.06-
Bleaching time	0	0-0.01	0.01-0.10	0.10-0.20	0.20-
pH	0	1.5-2.5 8.5-9.5	2.5-3.5 7.5-8.5	3.5-4.5 6.5-7.5	4.5-6.5
Maturing time	0	0-0.01	0.01-0.05	0.05-0.30	0.30-

The score of every fluorescent protein can be obtained shown as Figure1.

Because the fluorescent protein we wanted to choose had to be optimized in yeast and had to be given by parts, the following fluorescent proteins in Figure2 were all what we could choose from.

It shows that mCherry, mOrange and EYFP ranks top 3. However, after our pre-experiment, it was found that the signal of mOrange was too weak. The reason maybe was that we sequenced the part of mOrange and found that there were some mutations. Therefore, we finally chose mCherry(66.0336 ranking 12/478) and EYFP(34.9807 ranking 82/478) as our report genes.

The degradation of the fluorescent protein is also of great importance to our experiment. Therefore, the change of fluorescence intensity with time was measured, and here are the results.

To better explain the degradation of the fluorescent protein, we consulted a large number of documents^[2,3,4,5]. It shows that the degradation of the fluorescent protein is exponential.

The EYFP degradation curve can be described by function (1):
     

$$f(x) = a·e^{bx} + c·e^{dx} (1)$$

Coefficients (with 95% confidence bounds):

a = 133, b =-0.005066, c =-44.38, d =-0.02168

Goodness of fit:

SSE: 76.77

R-square: 0.9443

Adjusted R-square: 0.9363

RMSE: 1.912

The mCherry degradation curve also can be described by function (1), but the coefficients are different.

Coefficients (with 95% confidence bounds):

a = 613.8, b = -0.0003886, c = 0.0003207, d =0.06852

Goodness of fit:

SSE: 82.27

R-square: 0.9482

Adjusted R-square: 0.9404

RMSE: 2.028

The fitted degradation curve of mCherry approximates a straight line, so it can be simplified as a linear function:

$$ f(x) = p_1·x + p_2 (2)$$

Coefficients (with 95% confidence bounds):

$$ p_1 = -0.2287 , p_2 = 613.6$$

Goodness of fit:

SSE: 82.67

R-square: 0.9479

Adjusted R-square: 0.9455

RMSE: 1.938

It can be seen that the degradation rate of different fluorescent protein is very different, which made it clearer for us to know about the characteristics of the two fluorescent protein and offered important information to our experiment.

For the OD₆₀₀ values we got, we did some processing and modeling work. And here are our steps and results.

There were three groups in our experiment. They were blank control group, partial control group and experimental group. After getting all the data, first, we drew a histogram and a scattergram of time and maximum OD₆₀₀  values (Figure8, 9). These results were very instructive to experiments that these results told us the best measuring point and the best measuring interval.

From the beginning to the maximum OD₆₀₀ value, it fits the logistic model. The block effect of resource and environment for the growth of yeasts is reflected in the growth rate r, which makes r decrease with the increase in the number of yeasts x. Express r as a function r(x) of x, and take a simple and convenient linear reduction function r(x)=a+bx. In order to give a real meaning to the coefficients a and b in the growth rate function, we introduced two parameters:
(1)Intrinsic growth rate r r is the growth rate when x=0 (in theory);
(2)Population capacity x_mx_m  is the largest yeast amount that can be accommodated by resources and the When x=x_m, the quantity of yeasts is no longer increasing, that is r(x_m)=r+bx_m=0, then b=-r/x_m.
r and x_m values in our experiments are shown in the chart below.

&nbsp	YPD	SC	BY4741	D-THREE
R M			0.0164	0.0172
			0.8523	0.8034
	pABaC＋p1m	pABaC＋p1E	pABaC＋p2N	pABaC＋p1F
R M	-0.002364402	0.001746617	-0.002826764	-0.001905785
	0.402523944	0.508816901	0.424323944	0.542298592
	pCiRbS＋p1m	pCiRbS＋p2N	pCiRbS＋p1F	pbCiRS＋p1m
R M	-0.006367923	-0.007098618	-0.007176452	-0.007853975
	0.410507042	0.254873239	0.446169014	0.315098592
	pbCiRS＋p1E	pbCiRS＋p2N	pbCiRS＋p1F	pABaC+pCiRbS＋p1m
R M	-0.024143608	-0.012145451	0.002428334	-0.006280764
	0.413985915	0.458239437	0.270442254	0.337278873
	pABaC+pCiRbS+p1E	pABaC+pCiRbS＋p2N	pABaC+pCiRbS＋p1F	pABaC+pbCiRS＋p1m
R M	0.002305512	-0.00217225	0.002272595	0.002039534
	0.33171831	0.293661972	0.303701408	0.289346479
	pABaC+pbCiRS+P1e	pABaC+pbCiRS＋p2N	pABaC+pbCiRS＋p1F
R M	0.001894111	-0.003848457	-0.007151104
	0.301574648	0.345819718	0.329769014

The resulting growth rate function is $$r(x) = r(1 - {x \over x_m}) $$ Replacing intrinsic growth rate with r(x), get $${dx \over dt} = rx(1 - {x \over x_m}) , x(0) = x_0 (1)$$

Factor rx in the function shows the growth trend of yeast amount itself, while factor $$(1 - {x \over x_m}); reflects the block effects of resources and environment to the yeast quantity growth. Obviously, the bigger x is, the former factor is bigger and the latter factor is smaller. The growth of yeast amount is the result of the two factors.

Take x as the horizontal axis and dx/dt as the vertical axis, we obtained a parabola (Figure 1), when x = x_m/2, dx/dt reaches the maximum. As shown in Figure 10, dx/dt changes with the increasing x, and we can do the following analysis to the curve x(t).

Setting when t=0 x₀ &lt x_m/2, with the increase of t, dx/dt increases, so x is growing faster and faster and the curve x(t) is raised downward; dx/dt decreases when x_0 &gt x_m/2, x grows slower and slower and the curve x(t) is raised upward. x = x_m/2 is the inflection point of the curve. When x &rarr x_m , dx/dt &rarr 0, so x = x_m is the asymptote of x(t). From the above analysis, we can draw the figure as shown in Figure 11.

Actually, equation (1) can be solved by separation of a variable method as

$$x(t) = {x_m \over {1+({x_m \over x_0}-1)e^{-rt}}} (2)$$

By Analyzing our data, following results were obtained similar to the above. Figure12,13 shows the t-x curve and x-dx/dt curve of our blank control group, partial control group and experimental group. To make the results clearer, we enlarged two of them.